The following is a strengthening of juan's conclusion. We need to show that the logarithmic derivative of $$ f(q):=\left(\frac{1}{\sqrt{q}}-\sqrt{q}\right) \vartheta_2^2(q) $$ is negative. Multiplying the logarithmic derivative by $q$, this means that $$ -\frac{q}{1-q}-\sum_{n=1}^\infty\frac{8nq^{4n}}{1-q^{4n}}+\sum_{n=1}^\infty \frac{4nq^{2n}}{1+q^{2n}} < 0. $$ For $0 < q < 1$ we have $$ \sum_{n=1}^\infty\frac{8nq^{4n}}{1-q^{4n}} > \sum_{n=1}^\infty 8nq^{4n} = \frac{8q^4}{(1-q^4)^2} $$ and $$ \sum_{n=1}^\infty\frac{4nq^{2n}}{1+q^{2n}} < \sum_{n=1}^\infty 4nq^{2n} = \frac{4q^2}{(1-q^2)^2}, $$ hence it suffices to show that $$ -\frac{q}{1-q}-\frac{8q^4}{(1-q^4)^2}+\frac{4q^2}{(1-q^2)^2} < 0. $$ This holds for $0 < q < 0.37795$, hence in this range we are done.
Remark: One can generate larger ranges by keeping the first few terms in the sums, and estimating the tail similarly as above. The sums converge uniformly on any interval $[0,1-\epsilon]$, hence with a complementary argument as outlined by juan for $q\in[1-\epsilon,1]$, the above strategy should indeed work. All that is left now is numerical work, namely specifying the $\epsilon>0$ and the number of terms to be kept in the above sums.